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/**

* Write a program to find the node at which the intersection of two singly linked lists begins.

*

* For example, the following two linked lists:

*

* A: a1 → a2

* ↘

* c1 → c2 → c3

* ↗

* B: b1 → b2 → b3

* begin to intersect at node c1.

*

*

* Notes:

*

* If the two linked lists have no intersection at all, return null.

* The linked lists must retain their original structure after the function returns.

* You may assume there are no cycles anywhere in the entire linked structure.

* Your code should preferably run in O(n) time and use only O(1) memory.

*

*/

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) {

* val = x;

* next = null;

* }

* }

*/

public class IntersectionOfTwoLinkedLists160 {

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

int lA = 0;

ListNode tA = headA;

while (tA != null) {

lA++;

tA = tA.next;

}

int lB = 0;

ListNode tB = headB;

while (tB != null) {

lB++;

tB = tB.next;

}

if (lA < lB) {

for (int i=1; i<=(lB-lA); i++) headB = headB.next;

} else if (lB < lA) {

for (int i=1; i<=(lA-lB); i++) headA = headA.next;

}

while (headA != null && headB != null) {

if (headA == headB) {

return headA;

}

headA = headA.next;

headB = headB.next;

}

return null;

}

public ListNode getIntersectionNode2(ListNode headA, ListNode headB) {

if (headA == null || headB == null) return null;

ListNode a = headA;

ListNode b = headB;

while (a != null && b != null) {

if (a == b) return a;

a = a.next;

b = b.next;

}

if (a == null && b == null) return null;

if (a == null) {

a = headB;

} else {

b = headA;

}

while (a != null && b != null) {

a = a.next;

b = b.next;

}

if (a == null) {

a = headB;

} else {

b = headA;

}

while (a != null && b != null) {

if (a == b) return a;

a = a.next;

b = b.next;

}

return null;

}

/**

* https://leetcode.com/problems/intersection-of-two-linked-lists/discuss/49785/Java-solution-without-knowing-the-difference-in-len!

*

* You can prove that: say A length = a + c, B length = b + c, after

* switching pointer, pointer A will move another b + c steps, pointer B

* will move a + c more steps, since a + c + b + c = b + c + a + c, it

* does not matter what value c is. Pointer A and B must meet after

* a + c + b (b + c + a) steps. If c == 0, they meet at NULL.

*

*/

public ListNode getIntersectionNode3(ListNode headA, ListNode headB) {

//boundary check

if(headA == null || headB == null) return null;

ListNode a = headA;

ListNode b = headB;

//if a & b have different len, then we will stop the loop after second iteration

while( a != b){

//for the end of first iteration, we just reset the pointer to the head of another linkedlist

a = a == null? headB : a.next;

b = b == null? headA : b.next;

}

return a;

}

public ListNode getIntersectionNode4(ListNode headA, ListNode headB) {

if (headA == null || headB == null) return null;

ListNode pa = headA;

ListNode pb = headB;

while (pa != pb) {

pa = pa.next;

pb = pb.next;

if (pa == pb) return pa;

if (pa == null) pa = headB;

if (pb == null) pb = headA;

}

return pa;

}

}